Suppose $a$, $b$, $c$ and $d$ are integers satisfying: $a-b+c=5$, $b-c+d=6$, $c-d+a=3$, and $d-a+b=2$. What is the value of $a+b+c+d$?
Solution: Notice that in the system of equations, each variable is added twice and subtracted once. Thus, when we add all four equations together, the result is $a+b+c+d=5+6+3+2=\boxed{16}$.